Step by step please Thirty-five specimens of a new computer chip were tested for
ID: 3226280 • Letter: S
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Step by step please Thirty-five specimens of a new computer chip were tested for speed in certain in application, along with 35 specimens of chips with the old design. The average speed, in MHz, for the new chips was 495.6, and the standard deviation was 19.4. The average speed for the old chips was 481.2, and the standard deviation was 18.3. a. Can you conclude that the mean speed for the new chips is greater than that of the old chips? State the appropriate null and alternate hypotheses, and then find the P-value. b. A sample of 60 even older chips had an average speed of 391.2 MHz with a standard deviation of 17.2 MHz. Someone claims that the new chips average more than 100 MHz faster than these very old ones. Do the data provide convincing evidence for this claim? State the appropriate null and alternate hypotheses, and then find the P-value.Explanation / Answer
a.
H0 (the null hypothesis): That the difference, mA - mB , is equal to the chosen reference value (usually zero
· H1 (the alternative hypothesis): That mA - mB is not equal to the chosen reference value.
Here H1: mA>MB -----One tailed
Two-Sample T-Test and CI
Sample N Mean StDev SE Mean
1 35 495.6 19.4 3.3
2 35 481.2 18.3 3.1
Difference = (1) - (2)
Estimate for difference: 14.40
95% lower bound for difference: 6.88
T-Test of difference = 0 (vs >): T-Value = 3.19 P-Value = 0.001 DF = 68
Both use Pooled StDev = 18.8580
Conclusion : P-va;ue <0.05 When Minitab displays a p-value of 0.000, it means that the actual p-value is less than 0.0005. This p-value indicates that there is less than a 0.05% chance that you would have obtained your samples if mA - mB was actually 0
Two-Sample T-Test and CI
Sample N Mean StDev SE Mean
1 35 495.6 19.4 3.3
2 60 391.2 17.2 2.2
Difference = (1) - (2)
Estimate for difference: 104.40
95% lower bound for difference: 98.03
T-Test of difference = 100 (vs >): T-Value = 1.15 P-Value = 0.127 DF = 93
Both use Pooled StDev = 18.0354
P-value >0.05 we do not reject null hypotheis so we data do not provide convencing evidence for this claim,
Hope this will be helpful. Thanks and God Bless you:-)
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